Question

关于如何根据内部查询（循环查询）的结果解决查询问题，我遇到了一些问题。我已经尝试了所有我知道的方法，但它会中止while循环并中断。

它只是显示查询中的结果，它告诉我查询内的DOES有效，但它不会粘贴其信息，也不会粘贴其他元素（如信誉及其样式）。

$sql4 = mysqli_query($con, "SELECT * FROM as_sound_comments WHERE sound_id='$soundid'");

if(mysqli_num_rows($sql4) > 0){

    while($row = mysqli_fetch_assoc($sql4)) {
    echo "<div class='comment-content-full'>
            <div class='comment-info'>
                <table style='width: 98%'>
                    <td>
                        <table>
                            <td><p class='item-comment-info-name'><a class='blink' href='#'>";  
                                $comment_poster_id = $row['poster_id'];

                                $sql5 = mysqli_query($con, "SELECT * FROM as_account WHERE id='$comment_poster_id'");
                                $innerrow = mysqli_fetch_assoc($sql5);
                                $comment_poster_username = $innerrow['username'];
                                $comment_poster_reputation = $innerrow['reputationlevel'];

                                echo $comment_poster_username;
                                "</a></p></td>
                            <td><div class='item-comment-info-reputation nlgreen'>8</div></td>
                        </table>
                    </td>
                    <td style='width: 20%'>
                        <p class='item-comment-info-posttime'><abbr class='timeago' title='< ". $row['added'] . "'></abbr></p>
                    </td>
                </table>
            </div>

            <div class='comment-content-avatarandbox'> 
                <div class='comment-avatar'>
                    <img class='item-user-info-comment-avatar' src='images/test_avatar.jpg' alt='Avatar'>
                    <div class='arrow-left'> </div>
                    <div class='arrow-left-behind'> </div>
                </div>
                <div class='comment-box'>
                    <p class='comment-box-text'>" . $row['comments'] . "</p>
                </div>
            </div>
        </div>";

    }}else{ echo "sorry we didn't find no nothing";}

Answer 1

根据评论回答关闭问题。无论何时在查询循环中进行查询，都可以使用JOIN进行1次查询 -

$sql4 = mysqli_query($con, "SELECT * FROM as_sound_comments LEFT JOIN as_account ON as_account.id = as_sound_comments.poster_id WHERE as_sound_comments.sound_id='$soundid'");

使用结果查询查询

1 个答案: